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STRUCTURE OF ALUMINIUM ISOTOPES
By Prof. L. Kaliambos (Natural Philosopher in New Energy) 12 March 2019 Historically the discovery of the assumed uncharged neutron (1932) along with the invalid relativity (EXPERIMENTS REJECT RELATIVITY) led to the abandonment of the well-established electromagnetic laws, in favour of various contradicting nuclear theories, which could not lead to the nuclear structure. Under this physics crisis and using the charged UP and DOWN quarks , discovered by Gell-Mann and Zweig, I published my paper “Nuclear structure is governed by the fundamental laws of electromagnetism ”(2003), which led to my discovery of the new structure of protons and neutrons given by proton = + 5d + 4u = 288 quarks = mass of 1836.15 electrons neutron = + 4u + 8d = 288 quarks = mass of 1838.68 electrons The paper was also presented at a nuclear conference held at NCSR "Demokritos" (2002). In this photo I present the electromagnetic laws governing the nuclear structure, but a student of Einstein (Dr Th. Kalogeropoulos ) criticised my discovery of nuclear force and structure by believing that the nuclear structure is due to the invalid relativity. In fact, here one can see the 9 charged quarks in proton and the 12 ones in neutron able to give the charge distributions in nucleons for revealing the strong electromagnetic force for the nuclear binding in the correct nuclear structure by applying the laws of electromagnetism. Note that according to my discovery of the LAW OF ENERGY AND MASS the mass defect in the nuclear structure is due to the photon mass of the emitting dipolic photon presented at the international conference "Frontiers of fundamental physics" (1993) organised by the natural philosophers M. Barone and F. Selleri , who gave me an award including a disc of the atomic philosopher Democritus. Nevertheless today many physicist continue to apply not the well-established laws but the various fallacious nuclear structure models which lead to complications. Aluminium (Al) has 22 known isotopes from Al-21 to Al-42 and 4 known isomers. Only Al-27 (stable isotope) and Al-26 (radioactive isotope, t1/2 = 7.2 × 105 y) occur naturally, however Al-27 has a natural abundance of 99.9+ %. Other than Al-26, all radioisotopes have half-lives under 7 minutes, most under a second. ' ' WHY Al-27 WITH S = +5/2 IS A STABLE NUCLIDE After a careful analysis of the structure of atomic nuclei I discovered that the beta decay is due to the fact that in unstable nuclei there exist single horizontal pn bonds of weak binding energy leading to the beta decay. For example in my paper STRUCTURE AND BINDING OF H3 AND He3 using the diagram of the structure of the H3 one sees that it is unstable because the two neutrons make single np bonds, while the He3 is stable because the one neutron between the two protons makes two np bonds per neutron. On the other hand the pp repulsions of long range lead to the instability when we have a small number of pn bonds per nucleon. For understanding the stability of Al-27 with S = +5/2 you can read my STRUCTURE OF Mg-27, Al-27 AND Si-28 . Using the diagram of Al-27 we see that the 7 nucleons like the p11, n11, n13, p12, n12, p13 and n14 make strong vertical rectangles outside the parallelepiped (core) of 10 protons and 10 neutrons in which the 5 squares of opposite spins give S =+2. In other words the Al-17 is a stable nuclide with S = +5/2 because it has 17 nucleons of positive spins at the three horizontal planes like the +Hp1, the +Hp3 and the +Hp5, while at the -Hp2 and the -HP4 we observe 11 nucleons of negative spins. That is S = +17/2 -11/2 = +5/2 . ' DIAGRAM OF THE STABLE Al-27 WITH S = +5/2' ' ''' '' ' n10..........p10.....n14' ' p9...........n9 +HP5' ' p8.........n8.......p13' ' n7..........p7 -HP4' ' n6........p6.......n12' ' n11.........p5..........n5 +HP3' ' p4.........n4........p12' ' p11........n3..........p3 -HP2' ' n2........p2........n13' ' p1........n1 +HP1' ' ' ' NUCLEAR STRUCTURE OF ALUMINIUM NUCLIDES FOR A>27' After a careful analysis of all unstable nuclides of aluminium for A>27 I discovered that their structure is based on the structure of the Al-27 with S = +5/2. For example the structure of the Al-28 with S =+3 has one extra neutron of positive spin which makes a single horizontal bond with a proton . That is S =+5/2 +1/2 = +6/2 =+3 Moreover in the Al-41 with S = +3/2 having 14 extra neutrons than those of Al-27 we conclude that it has two extra neutrons of negative spin and 12 extra neutrons of opposite spins giving S = 0. That is S = +5/2 -1/2 -1/2 + 0 = +3/2. However in the case of the Al-34 with S = -4 we conclude that the nucleons of the Al-27 change their spins. For example in this case we have -HP1, +HP2, -Hp3. +Hp4 and -Hp5 with S = -5/2. Since the Al-34 has 7 more extra neutrons than those of Al-27 we conclude that it has 3 extra neutrons of negative spins and 4 extra neutrons of opposite spins giving S = 0. That is S = -5/2 -1/2 -1/2- 1/2 + 0 = -8/2 = -4. NUCLEAR STRUCTURE OF Al-26, Al-25, Al-24, Al-23, Al-22 AND Al-21 After a careful analysis of the unstable nuclides with A<27 I discovered that their structure is based on the parallelepiped ( core) of 10 protons and 10 neutrons giving S =+2 due to the fifth horizontal square, the p9n9p10n10. In the absence of a neutron we get the Al-26 with S = +5. Here we see that in the absence of one neutron outside the parallelepiped of S =+2 we have 3 protons and 3 neutrons . In other words we have 3 deuterons of S = +3, because they makehorizontal bonds on the three sides of the parallelepiped at the +HP3. In other words at the third horizontal plane we have 10 nucleons of positive spins which give the total spin S = +5. ' '''Then in the absence of two neutrons we observe that outside the parallelepiped of S =+2 there are 3 protons and 2 neutrons giving the Al-25 with S = +5/2. In the following diagram of Al-25 you see that the p12, n12 and p13 make vertical strong bonds while the p11 and n11 make a horizontal square with horizontal bonds.. ' ' ' DIAGRAM OF Al-25 WITH S =+5/2''' ' n10........p10..... n14' ' p9...........n9 +HP5' ' p8.........n8.......p13' ' n7..........p7 -HP4' ' p11.......n6........p6.........n12' ' n11.........p5..........n5 +HP3' ' p4.........n4........p12' ' n3..........p3 -HP2' ' n2........p2' ' p1........n1 +HP1' However in the absence of thee neutrons we get the structure of Al-24 with S = +4. In this case we discovered that two absent neutrons from the corners of the parallelepiped like the n1 and the n10 reduce the number of bonds. Thus outside the parallelepiped we have 3 protons and 3 neutrons or 3 deuterons of S =+3 which make horizontal bonds a t the +HP3 as in the case of Al-26. But here in the absence of two neutrons the spin of the parallelepiped is S = +1 given by ' '+HP1= 3nucleons of positive spins -HP2 = 4 nucleons of negative spins +Hp3 = 4 nucleons of positive spins -HP4 = 4 nucleons of negative spins +Hp5 = 3 nucleons of positive spins ' '''Therefore the total spin of Al-26 with S = +4 is due to the summation of the S = +1 of the parallelepiped and the S = +3 of the three deuterons. Now using the same method we see that the structure of Al-23 with S = +5/2 is due to the fact that outside the parallelepiped with S = +1 there are 3 protons and 2 neutrons giving S= = +3/2 . In this case the two protons along with the one neutron make two vertical rectangles with S = +1/2, while the one proton and the one neutron make a deuteron with S' =+1. ' That is S = +1 +1/2 +1 = +5/2' . ' In the Al-22 of S =+3 the three protons and the one neutron existing outside the parallelepiped of S = +1 have positive spins with S = +2 . Whereas in the Al-21 with S = +1/2 we have two protons existing outside the parallelepiped with negative spins and one proton with positive spin. That is S = +1 -1/2 -1/2 +1/2 = +1/2. . ' ''' Category:Fundamental physics concepts